# Page 149 Location: Chapter 14, bonus exercise 1, proposition (d)

## It is

(d) If $Px(Pyz)$ sings on all days, so does $P(Pxy)(Pyz)$.

## It should disappear

[no sentence (d) to prove at all]

## Short explanation

From the first three of Byrd's laws, proposition (d) does not follow. (The rest of the exercise proceeds without problems, though.)

In the terms of Problem 1 of the Chapter, the particular arrangement

x \notin S~~,~~~y \in S~~,~~~z \notin S

acts as a counterexample, making it impossible to derive the fact that $P(Pxy)(Pyz) \in S$ from $Px(Pyz)\in S$ using the first three laws of Problem 1 alone.

In terms of the usual Aristotelian logic (i.e. under the isomorphism presented in "Discussion" on page 145), the following truth table makes the counterexample explicit:

$x$ $y$ $z$ $y\to z$ $z\to y$ $(x\to y)\to(y\to z)$ $x\to(y\to z)$ $[x\to(y\to z)]\to[(x\to y)\to(y\to z)]$
T T T T T T T T
T T F F T F F T
T F T T F T T T
T F F T F T T T
F T T T T T T T
F T F F T F T F
F F T T T T T T
F F F T T T T T